sin(4x) − sin(3x) = sin(2x) find x ?


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Question Number 163822 by mkam last updated on 11/Jan/22

$s i n (4 x) - s i n (3 x) = s i n (2 x) f i n d x ?$
	Answered by cortano1 last updated on 11/Jan/22

	Commented by mr W last updated on 11/Jan/22

	$p l e a s e r e c h e c k . i t h i n k a n s w e r i s$ $w r o n g .$
	Commented by cortano1 last updated on 11/Jan/22

	$w h a t w r o n g!$
	Commented by mr W last updated on 11/Jan/22

	$x = \pm \cos^{- 1} \frac{2 - \sqrt[3]{10} - \sqrt[3]{100}}{6} + 2 k π {d o e s n}^{'} t$ $f u l f i l l t h e e q u a t i o n .$
	Answered by mr W last updated on 11/Jan/22
	$sin 4x−sin 2x=sin 3x sin (3x+x)−sin (3x−x)=sin 3x 2 cos 3x sin x=sin 3x 2(4 cos^3 x−3 cos x)sin x=sin x (3−4 sin^2 x) ⇒sin x=0 ⇒x=nπ or 8 cos^3 x−6 cos x=3−4 sin^2 x 8 cos^3 x−4 cos^2 x−6 cos x+1=0 cos^3 x−(1/2) cos^2 x−(3/4) cos x+(1/8)=0 let cos x=t+(1/6) t^3 −((5t)/6)−(1/(108))=0 t=((√(10))/3) sin (((2kπ)/3)−(1/3)sin^(−1) ((√(10))/(100))) cos x=(1/6)+((√(10))/3) sin (((2kπ)/3)−(1/3)sin^(−1) ((√(10))/(100))) such that −1≤cos x≤1, ⇒ k=0, 2 cos x=(1/6)−((√(10))/3) sin ((1/3)sin^(−1) ((√(10))/(100))) >0 cos x=(1/6)−((√(10))/3) sin ((π/3)−(1/3)sin^(−1) ((√(10))/(100))) <0 ⇒x=2nπ±cos^(−1) {(1/6)−((√(10))/3) sin ((1/3)sin^(−1) ((√(10))/(100)))} ⇒x=(2n+1)π±cos^(−1) {((√(10))/3) sin ((π/3)−(1/3)sin^(−1) ((√(10))/(100)))−(1/6)}$
	$\sin 4 x - \sin 2 x = \sin 3 x$ $\sin (3 x + x) - \sin (3 x - x) = \sin 3 x$ $2 \cos 3 x \sin x = \sin 3 x$ $2 (4 \cos^{3} x - 3 \cos x) \sin x = \sin x (3 - 4 \sin^{2} x)$ $\Rightarrow \sin x = 0 \Rightarrow x = n π$ $o r$ $8 \cos^{3} x - 6 \cos x = 3 - 4 \sin^{2} x$ $8 \cos^{3} x - 4 \cos^{2} x - 6 \cos x + 1 = 0$ $\cos^{3} x - \frac{1}{2} \cos^{2} x - \frac{3}{4} \cos x + \frac{1}{8} = 0$ $l e t \cos x = t + \frac{1}{6}$ $t^{3} - \frac{5 t}{6} - \frac{1}{108} = 0$ $t = \frac{\sqrt{10}}{3} \sin (\frac{2 k π}{3} - \frac{1}{3} \sin^{- 1} \frac{\sqrt{10}}{100})$ $\cos x = \frac{1}{6} + \frac{\sqrt{10}}{3} \sin (\frac{2 k π}{3} - \frac{1}{3} \sin^{- 1} \frac{\sqrt{10}}{100})$ $s u c h t h a t - 1 ⩽ \cos x ⩽ 1, \Rightarrow k = 0, 2$ $\cos x = \frac{1}{6} - \frac{\sqrt{10}}{3} \sin (\frac{1}{3} \sin^{- 1} \frac{\sqrt{10}}{100}) > 0$ $\cos x = \frac{1}{6} - \frac{\sqrt{10}}{3} \sin (\frac{π}{3} - \frac{1}{3} \sin^{- 1} \frac{\sqrt{10}}{100}) < 0$ $\Rightarrow x = 2 n π \pm \cos^{- 1} {\frac{1}{6} - \frac{\sqrt{10}}{3} \sin (\frac{1}{3} \sin^{- 1} \frac{\sqrt{10}}{100})}$ $\Rightarrow x = (2 n + 1) π \pm \cos^{- 1} {\frac{\sqrt{10}}{3} \sin (\frac{π}{3} - \frac{1}{3} \sin^{- 1} \frac{\sqrt{10}}{100}) - \frac{1}{6}}$
	Commented by peter frank last updated on 11/Jan/22

	$thank you$
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